Problem: $\begin{aligned}4x+4y-3z-t&=12 \\-10x-y+3t&=14\end{aligned}$ Which of the following matrices represents this system? Choose 1 answer: Choose 1 answer: (Choice A) A $ \left[\begin{array} {ccc} 4 & 4 & -3 & -1 & 12 \\ -10 & -1 & 0 & 3 & 14 \end{array} \right]$ (Choice B) B $ \left[\begin{array} {ccc} 4 & 4 & -3 & -1 & -12 \\ -10 & -1 & 1 & 3 & -14 \end{array} \right]$ (Choice C) C $ \left[\begin{array} {ccc} 4 & 4 & -3 & -1 & -12 \\ -10 & -1 & -1 & 3 & -14 \end{array} \right]$ (Choice D) D $ \left[\begin{array} {ccc} 4 & 4 & -3 & -1 & 12 \\ -10 & -1 & 3 & 1 & 14 \end{array} \right]$
Answer: Background We can express a system of equations in an augmented matrix in which each row represents one equation. To do this, we write the coefficients of each variable in its own column, and the constants on the right side of the equations in the last column. For example, if we are given a system of equations in two variables, then we can express this as an augmented matrix as follows. $\begin{aligned}ax+by&=m \\cx+dy&=n\end{aligned}$ $ \rightarrow$ $ \left[\begin{array} {ccc} a & b & m \\ c & d & n \end{array} \right]$ Representing the system of equations as a matrix We are given the system of equations: $\begin{aligned}4x+4y-3z-t&=12 \\-10x-y+3t&=14\end{aligned}$ First, let's rewrite this system to show the coefficients of each variable. $\begin{aligned}{4}x+{4}y+({-3})z+({-1})t&={12} \\({-10})x+({-1})y+{0}z+{3}t&={14}\end{aligned}$ [Why does z appear in the second equation here, although it wasn't there originally?] Now, we can write our augmented matrix. $\left[\begin{array} {ccc} 4 & 4 & -3 & -1 & 12 \\ -10 & -1 & 0 & 3 & 14 \end{array} \right]$ Summary This is the correct augmented matrix. $\left[\begin{array} {ccc} 4 & 4 & -3 & -1 & 12 \\ -10 & -1 & 0 & 3 & 14 \end{array} \right]$